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The Bell basis is the two-qubit case of the generalized Bell basis. Its four vectors are the Bell states, and a Bell-basis measurement projects onto them.

Two initial maximally entangled pairs $A{B}_1$ and $B_{2}C$ can be turned into an entangled pair $AC$ by a generalized Bell basis measurement on $B_1{B}_2$, followed by a local correction at an endpoint using its classical label. This is qudit teleportation of $B_1$, including its correlations with $A$, to $C$. The initial two pairs are consumed. Without the classical label, averaging the conditional endpoint states need not leave entanglement.

A generalized Bell state is one member of a generalized Bell basis. Its reduced density matrices are both $I/n$, so it is a maximally entangled state. For $n=2$ these are the four Bell states.

Take the labels $r,s\in \{0,\dots ,n-1\}$, with basis indices read modulo $n$. There is a missing hypothesis in the printed claim: the root of unity must be primitive. For example, when $n=2$ and $\omega =1$, the two labels $(0,0)$ and $(1,0)$ produce the identical state $(|00⟩+|11⟩)/\sqrt{2}$, so they are not orthogonal. More generally, if the order of $\omega$ is $d<n$, changing $r$ to $r+d$ repeats a state. The orthonormality claim is false for a nonprimitive root.

For the intended result, choose a primitive root of unity, for example $\omega =e^{2\pi i/n}$. Since $|\omega |=1$, expansion in the orthonormal basis givesIf $r=r^{\prime }$, the sum is $n$. If $r\ne r^{\prime }$ modulo $n$, primitiveness gives $q={\omega }^{r^{\prime }-r}\ne 1$ while $q^n=1$, and the finite geometric series is $(1-q^n)/(1-q)=0$. ThusThere are $n^2$ vectors in an $n^2$-dimensional Hilbert space, so they form the generalized Bell basis. Each has equal Schmidt coefficients $1/\sqrt{n}$ and is a maximally entangled state.

For qudit teleportation, let $C$ be Alice's unknown input, write $|\chi {⟩}_C={\sum }_j{\alpha }_j|j{⟩}_C$, and let Alice and Bob share $|\Phi {⟩}_{AB}=n^{-1/2}{\sum }_k|k{⟩}_A|k{⟩}_B$. Alice holds $C,A$ and Bob holds $B$. Define the qudit shift and phase operatorsBoth are unitary. Alice makes a joint projective measurement of $C,A$ in the generalized Bell basis, with projectors $|{\psi }_{rs}⟩⟨{\psi }_{rs}|$. Completeness follows from the orthonormality just proved.

For outcome $(r,s)$, contracting the corresponding bra with the joint input and resource gives Bob's unnormalized state:The Born rule therefore gives $p_{rs}=1/n^2$, since $X^s{Z}^{-r}$ preserves the input norm. Conditional on this outcome, Bob holds $X^s{Z}^{-r}|\chi ⟩$. Alice sends him the classical label $(r,s)$, and he applies its inverse in the displayed order:The corrected receiver state is exactly the unknown input state for every outcome. Alice needs no knowledge of the amplitudes. The label has $n^2$ possibilities and can be encoded in $\lceil {\log }_2(n^2)\rceil$ classical bits; when $n$ is a power of two this is exactly $2{\log }_{2}n$. The protocol consumes the shared maximally entangled state and the original carrier is measured, so it does not retain an additional copy of the unknown state.

Equivalently, the complete joint quantum state has the teleportation expansionIt follows from the contraction calculation and completeness of the generalized Bell basis. After correction the branch operators are all $I/n$, so their sum defines the identity quantum channel: ${\sum }_{r,s}(I/n)\rho (I/n)=\rho$. Acting on the input part of a joint state with a reference gives the same identity, which proves teleportation as an identity channel on a reference and extends the protocol to mixed inputs.

Before receiving the label, Bob instead has the outcome-averaged stateTo check the equality, averaging over $r$ kills the off-diagonal matrix entries by the same root-of-unity sum; averaging the cyclic shifts $s$ makes every diagonal entry the total trace divided by $n$. This is the Heisenberg-Weyl twirling channel. Thus the classical message is necessary to recover the input, consistently with the no-communication theorem derived in Question 1.

Label an orthonormal basis of the $n$-dimensional Hilbert space by $j=0,\dots ,n-1$, with indices modulo $n$. LetThese are the shift and phase Heisenberg-Weyl operators. Alice holds the input $S$ and one half $A$ of the shared maximally entangled stateShe measures $SA$ in the generalized Bell basisIts inner products areThere are $n^2$ orthonormal vectors in an $n^2$-dimensional space, so this is a complete projective measurement.

For an unknown normalized state $|u⟩={\sum }_j{u}_j|j⟩$, its contraction with measurement outcome $(p,q)$ gives Bob's unnormalized stateEach outcome has probability $1/n^2$, independently of the input. Alice communicates the two $n$-valued labels, and Bob applies the ordered inverseThe order matters because shift and phase operators do not generally commute. Normalizing the branch recovers the unknown state exactly, and all outcomes are corrected, so success is deterministic. The quantum resource is one maximally entangled pair of local dimension $n$. The classical resource is an $n^2$-valued message, with information content $2{\log }_{2}n$ bits; $\lceil {\log }_2(n^2)\rceil$ ordinary bits suffice in fixed-length binary encoding.

This is faithful qudit teleportation even for an input entangled with a reference $R$. Expand a joint pure state as ${\sum }_j|j{⟩}_S|r_j{⟩}_R$, without requiring the reference vectors to be orthogonal. The same contraction and correction act on each system basis vector, producing $n^{-1}{\sum }_j|j{⟩}_B|r_j{⟩}_R$. Thus the whole joint state, not just a reduced state, is preserved. Linearity extends the conclusion to mixed inputs. This is teleportation as an identity channel on a reference.

The root $\omega$ must be a primitive $n$th root of unity, for example $e^{2\pi i/n}$. The printed statement merely says $n$th root; that is insufficient. With $n=2$ and $\omega =1$, the states with $r=0$ and $r=1$ coincide. More generally, a root of order $d<n$ repeats the $r$ labels modulo $d$. For $n=1$ the construction is trivial.

For a primitive root of unity, the generalized Bell basis obeysThe last equality follows from the finite geometric series: a nonzero exponent difference modulo $n$ has sum zero, while zero difference has sum $n$. There are $n^2$ orthonormal vectors in an $n^2$-dimensional Hilbert space, so they form a complete basis.

For qudit teleportation, Alice's input is $|\chi {⟩}_U={\sum }_j{c}_j|j⟩$. Alice and Bob share $|{\psi }_{00}{⟩}_{aB}=n^{-1/2}{\sum }_k|kk⟩$. Define the qudit shift and phase operators by $X|j⟩=|j+1⟩$ and $Z|j⟩={\omega }^j|j⟩$. Alice measures $Ua$ in the generalized Bell basis. On outcome $(r,s)$, Bob's unnormalized state isAll outcomes have probability $1/n^2$. Alice communicates $(r,s)$ and Bob applieswhich restores $|\chi ⟩$ exactly. The protocol consumes one maximally entangled pair of $n$-level systems, a local $n^2$-outcome measurement, and a classical message with $n^2$ possibilities. For a fixed-length binary encoding, $\lceil {\log }_2{n}^2\rceil$ bits suffice. No measurement depends on the unknown amplitudes.

Bob uses qudit teleportation to teleport his carrier $B_1$ to Clare through the pair $B_{2}C$. This transfers Bob's entanglement with Alice to Clare. Explicitly, Bob performs a generalized Bell basis measurement on $B_1{B}_2$. Contraction of the two initial pairs givesThe probability is $1/n^2$ for each outcome. Bob sends $(r,s)$ to Clare, who applies $Z_C^r{X}_C^{-s}$. Alice and Clare then share $|{\psi }_{00}{⟩}_{AC}$ with certainty. This is entanglement swapping; neither an additional entangled pair nor a quantum transmission during the protocol is required.

The two initial pairs are consumed, and Bob's measured carriers cease to be entangled with $AC$. Without Bob's classical record, averaging the possible generalized Bell states leaves $AC$ maximally mixed. Thus the conditional entanglement swapping does not supply faster-than-light communication.

A sender and receiver share $n^{-1/2}{\sum }_j|jj⟩$. The sender measures the input and their resource half in the generalized Bell basis. Outcome $(r,s)$ gives the receiver $X^s{Z}^{-r}|\chi ⟩/n$, with probability $1/n^2$. After classical communication of the outcome, the correction $Z^r{X}^{-s}$ restores the unknown quantum state. The same calculation preserves correlations with a reference, as in teleportation as an identity channel on a reference.